Question

Sukie interviewed 125 employees at her company and discovered that 21 of them planned to take an extended vacation next year. what is the 95% confidence interval for this population proportion?

O 0.10 to 0.23
O 0.16 to 0.17
O 0.11 to 0.16
O 0.11 to 0.21

Answer 1

The 95% confidence interval for the sample proportion of employees planning to take an extended vacation is approximately 0.10 to 0.23. This was calculated using the sample proportion and applying the z-score for the 95% confidence level. Option 1 is the correct answer.

Step-by-step explanation:

To calculate the 95% confidence interval for a population proportion based on the sample, we use the formula:

Confidence Interval = π ± Z * √(π(1 - π) / n)

Where π is the sample proportion, Z is the z-score corresponding to the confidence level (in this case, 1.96 for 95%), and n is the sample size.

Let's calculate the sample proportion (p) first:

p = number of favorable responses / total responses

p = 21 / 125

p = 0.168

Now, we use the standard error of the sample proportion formula:

Standard error (SE) = √(p(1 - p) / n)

SE = √(0.168(1 - 0.168) / 125)

SE = 0.0335

Now we can calculate the confidence interval:

Confidence Interval = 0.168 ± 1.96 * 0.0335

Upper Limit = 0.168 + (1.96 * 0.0335) = 0.235

Lower Limit = 0.168 - (1.96 * 0.0335) = 0.101

Therefore, the 95% confidence interval is approximately 0.10 to 0.23.

The correct answer from the options provided is 0.10 to 0.23.